PHP call array value

Question

First I have these variables:

$asd=$current[0]->icon['data']; // -it will print  /ig/images/weather/mostly_cloudy.gif
$icons["/ig/images/weather/rain.gif"][0]; // it will print rain.gif

I want to see rain.gif output. I'm trying to do this.

echo $icons[$asd][0]; // I will get error. 

I want to get this output with variables.

$icons["/ig/images/weather/rain.gif"][0]; // it will print rain.gif

Is the code wrong?

echo $icons[$asd][0];

Please help.

thanks evryone. sorry for my english.

how can I do this.

Answer 1

I think you're looking for basename.

echo basename('/ig/images/weather/mostly_cloudy.gif'); // output: mostly_cloudy.gif');
echo basename('/ig/images/weather/rain.gif');          // output: rain.gif

Assuming that $asd (in your example) is the path to the image and you're just looking for the file name portion).

Though your question is not 100% clear.